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JEE Main

Explore popular questions from Dual Nature of Matter and Radiation for JEE Main. This collection covers Dual Nature of Matter and Radiation previous year JEE Main questions hand picked by experienced teachers.

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Dual Nature of Matter and Radiation

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Q 1. Radiations of frequency {tex}v{/tex} are incident on a photosensitive metal. The maximum K.E. of the photoelectrons is {tex}E{/tex}. When the frequency of the incident radiation is doubled, what is the maximum kinetic energy of the photoelectrons?

A

{tex}2E{/tex}

B

{tex}4E{/tex}

{tex}E + hv{/tex}

D

{tex}E – hv{/tex}

Explanation

Using Einstein’s photoelectric equation {tex}hv – \phi = E{/tex}

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Q 2. Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure)

In an experiment, electrons are made to pass through a narrow slit of width {tex} d {/tex} comparable to their de Broglie wavelength. They are detected on a screen at a distance {tex} D {/tex} from the slit.

Which of the following graphs can be expected to represent the number of electrons {tex} N {/tex} detected as a function of the detector position {tex} y ( y = 0 {/tex} corresponds to the middle of the slit)?

B

C

D

Explanation



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Q 3. If {tex} g _ { E } {/tex} and {tex} g _ { M } {/tex} are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratio \[ \frac { \text { electronic charge on the moon } } { \text { electronic charge on the earth } } \text { to be } \]

A

{tex} g _ { M } / g _ { E } {/tex}

1

C

0

D

{tex} g _ { E } / g _ { M } {/tex}

Explanation

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Q 4. Photon of frequency {tex} v {/tex} has a momentum associated with it. If {tex} c {/tex} is the velocity of light, the momentum is

{tex} h \nu / c {/tex}

B

{tex} \nu / c {/tex}

C

{tex} h\nu c {/tex}

D

{tex} h v / c ^ { 2 } {/tex}

Explanation


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Q 5. The anode voltage of a photocell is kept fixed. The wavelength {tex} \lambda {/tex} of the light falling on the cathode is gradually changed. The plate current {tex} I {/tex} of the photocell varies as follows

A

B

D

Explanation


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Q 6. Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure)

If a strong diffraction peak is observed when electrons are incident at an angle {tex} i {/tex} from the normal to the crystal planes with distance {tex} d {/tex} between them (see figure), de Broglie wavelength {tex} \lambda _ { d B } {/tex} of electrons can be calculated by the relationship {tex} ( n {/tex} is an integer)

A

{tex} d \cos i = n \lambda _ { d B } {/tex}

B

{tex} d \sin i = n \lambda _ { d B } {/tex}

{tex} 2 d \cos i = n \lambda _ { d B } {/tex}

D

{tex} 2 d \sin i = n \lambda _ { d B } {/tex}

Explanation



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Q 7. The anode voltage of a photocell is kept fixed. The wavelength {tex} \lambda {/tex} of the light falling on the cathode is gradually changed. The plate current {tex} I {/tex} of the photocell varies as follows

B

C

D

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Q 8. This question has Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement 1: Davisson - Germer experiment established the wave nature of electrons.
Statement 2: If electrons have wave nature, they can interfere and show diffraction.

A

Statement 1 is true, Statement 2 is false.

Statement 1 is true, Statement 2 is true, Statement 2 is the correct explanation for Statement 1.

C

Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1.

D

Statement 1 is false, Statement 2 is true.

Explanation



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Q 9. This question has Statement-l and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements :
Statement-1: A metallic surface is irradiated by a monochromatic light of frequency {tex} v > v _ { 0 } {/tex} (the threshold frequency). The maximum kinetic energy and the stopping potential are {tex} K _ { \max } {/tex} and {tex} V _ { 0 } {/tex} respectively. If the frequency incident on the surface is doubled, both the {tex} K _ { \max } {/tex} and {tex} V _ { 0 } {/tex} are also doubled.
Statement-2: The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.

A

Statement-1 is true, statement-2 is false.

B

Statement-l is true, Statement-2 is true, Statement-2 is the correct explanation of Statement- {tex} 1 . {/tex}

C

Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of Statement- {tex} 1 . {/tex}

Statement-1 is false, Statement-2 is true.

Explanation



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Q 10. If a source of power {tex} 4 \mathrm { kW } {/tex} produces {tex} 10 ^ { 20 } {/tex} photons/second, the radiation belongs to a part of the spectrum called

A

{tex} \gamma {/tex} -rays

{tex} \mathrm { X } {/tex} -rays

C

ultraviolet rays

D

microwaves

Explanation

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Q 11. Statement-1: When ultraviolet light is incident on a photocell, its stopping potential is {tex} V _ { 0 } {/tex} and the maximum kinetic energy of the photoelectrons is {tex} K _ { \max } . {/tex} When the ultraviolet light is replaced by X-rays, both {tex} V _ { 0 } {/tex} and {tex} K _ { \max } {/tex} increase.
Statement-2: Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

Statement-1 is true, Statement-2 is false.

B

Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-I.

C

Statement-l is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.

D

Statement-1 is false, Statement-2 is true.

Explanation



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Q 12. Sodium and copper have work functions {tex} 2.3 \mathrm { eV } {/tex} and {tex} 4.5 \mathrm { eV } {/tex} respectively. Then the ratio of the wavelengths is nearest to

A

1:2

B

4:1

2:1

D

1:4

Explanation

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Q 13. The time taken by a photoelectron to come out after the photon strikes is approximately

A

{tex} 10 ^ { - 1 } s {/tex}

B

{tex} 10 ^ { - 4 } s {/tex}

{tex} 10 ^ { - 10 } \mathrm { s } {/tex}

D

{tex} 10 ^ { - 16 } \mathrm { s } {/tex}

Explanation

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Q 14. Threshold wavelength for a metal having work function {tex}W_0{/tex} is {tex}X{/tex}. What is the threshold wavelength for the metal having work function {tex}2W_0{/tex}?

A

{tex}4\lambda{/tex}

B

{tex}2\lambda{/tex}

{tex}\frac{\lambda}{2}{/tex}

D

{tex}\frac{\lambda}{4}{/tex}

Explanation

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Q 15. The work function of a substance is {tex} 4.0 \mathrm { eV } . {/tex} The longest wavelength of light that can cause photoclectron emission from this substance is approximately

A

{tex} 540 \mathrm { nm } {/tex}

B

{tex} 400 \mathrm { nm } {/tex}

{tex} 310 \mathrm { nm } {/tex}

D

{tex} 220 \mathrm { nm } {/tex}

Explanation

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Q 16. Two identical photocathode receive light of frequency ν1 and ν2. If the velocities of the photoelectrons (of mass m) coming out are v1 and v2, respectively, then-

A

v1 - v2 =

=

C

v1 + v2 =

D

=

Explanation

1 = hν0 +
2 = hν0 +
h(ν1 - ν2) = m ()

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Q 17. Masses of two isobars 29Cu64 and 30Zn64 are 63.9298 u and 63.9292 u respectively. It can be concluded from these data that

A

Both the isobars are stable

B

Zn64 is radioactive decaying to Cu64 through β - decay.

C

Cu64is radioactive decaying to Zn64 through γ-decay.

Cu64 is radioactive decaying to Zn64 through β-decay.

Explanation

Cu64 is radioactive after β-decay its atomic number increases by one and its mass number doesn't change.
it decays to Zn64

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Q 18. An α - particle after passing through a potential difference of V-volts collides with a nucleus. If the atomic number of the nucleus is Z then the distance of closest approach of α-particle to the nucleus will be-

14.4

B

14.4 m

C

14.4 cm

D

All of these

Explanation

K.E. = P.E. = qV
2eV = = qV
d =
d =
d = 14.4 × 10-10 m

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Q 19. The order of energies of energy levels A, B and C is EA< EB< EC. If the wavelength corresponding to transition C → B, B → A and C → A are λ1, λ2 and λ3 respectively, then which of the following relation is correct?

A

λ1 + λ2 + λ3 = 0

B

λ32 = λ12 + λ22

C

λ3 = λ1 + λ2

λ3 =

Explanation

ECA = ECB + EBA

=
λ3 =

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Q 20. Which the following series fall in the visible range of electromagnetic spectrum?

14.4

B

14.4 m

C

14.4 cm

D

All of these

Explanation

K.E. = P.E. = qV
2eV = = qV
d =
d =
d = 14.4 × 10-10 m

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Q 21. The ratio of minimum to maximum wavelength in Balmer series is-

B

C

D

Explanation

For Balmer series = R
put n = ∞, =
put n = 3, = R
= =

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Q 22. The ratio of minimum wavelength of Lyman and Balmer series will be-

A

1.25

0.25

C

5

D

10

Explanation

For lyman;
= R = R
λmin =
For Balmer;
= R =
λmin =
= 0.25

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Q 23. The mean lives of a radioactive substance are 1620 years and 405 years for α-emission and β-emission respectively. Find the time during which three-fourth of a sample will decay if it is decaying both by α-emission and β-emission simultaneously -

A

249 years

449 years

C

133 years

D

99 years

Explanation

The decay constant λ is the reciprocal of the mean life τ.
Thus, λα = per year
and λβ = per year
Total decay constant, λ = λα + λβ or λ
= per year
We know that N = N0 e-λt
When th part of the sample has disintegrated, N = N0/4
= N0e-λt
or eλt = 4
Taking logarithm of both sides, we get
λt = loge 4
or t = loge 22 = loge 2
= 2 × 324 × 0.693 = 449 year

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Q 24. Kα wavelength of an unknown element is .0709 nm. Identify the element-

A

Co

B

Cu

C

Mn

Mo

Explanation

f = = = 4.23 × 1018 Hz
From Moseley's law Z = 1 +
Z = 42 corresponds to Mo.

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Q 25. Consider the nuclear reaction X200→ A 110 + B80. If the binding energy per nucleon for X, A and B are 7.4 MeV, 8.2 MeV and 8.1 MeV respectively, then the energy released in the reaction -

70 MeV

B

200 MeV

C

190 MeV

D

10 MeV

Explanation

For X : energy = 200 × 7.4 = 1480 MeV
For A : energy = 110 × 8.2 = 902 MeV
For B : energy = 80 × 8.1 = 648 MeV
Energy released = (902 + 648) - 1480
= 1550 - 1480
= 70 MeV